d and f Block Elements

Species having same number of electrons are isoelectronic

$\left.\begin{array}{l}\mathrm{Tb} \longrightarrow 65 \mathrm{~Tb}^{+2} \longrightarrow 63 \text { electrons } \\ \mathrm{Tm} \longrightarrow 69 ~\mathrm{Tm}^{+4} \longrightarrow 65 \text { electrons }\end{array}\right\}$ not isoelectronic

$\left.\begin{array}{c}\mathrm{Sm}^{+2} \longrightarrow 60 \text { electrons } \\ \mathrm{Er}^{+3} \longrightarrow 65 \text { electrons }\end{array}\right\}$ not isoelectronic

(6) Diamagnetic - No unpaired electrons. All the electrons are paired. A) $\mathrm{La}^{3+}, \mathrm{Ce}^{4+}$. Element $L a$ election configuration is $[\mathrm{Xe}] 5 \mathrm{~d}^{\prime} 6 \mathrm{~s}^2$ For $\mathrm{La}^{3+}$, three valence electrons are removed So, the electron configuration of $\mathrm{La}^{3+}$ is $\left[\mathrm{Xe}\right]$ The configuration [ $x_e$ ] is the noble gas configuration and all the elections are paired. So, $\mathrm{La}^{3+}$ is diamagnetic. Element $\mathrm{Ce}_{\text { }}$ election configuration is $\left[x_e\right] 4 f^{\prime} 5 d^{\prime} 6 s^2$ For $\mathrm{Ce} ^{4+}$, for valence elections are removed. so, the electron configuration of $\mathrm{Ce}^{4+}$ is $\left[\mathrm{Xe}\right]$. The origination $\left[X_e\right]$ is the noble gas configuration and all the electrons are paired. So, $Ce^{4+}$ is diamagnetic. So, It is a pair of a diamagnetic ions. B) $\mathrm{Yb}^{2+}, \mathrm{Lu}^{3+}$ Element Yb electron configuration is : $\left[x_e\right] 4 f^{14} 6 s^2$ For $y b^{2+}$, two valence electrons are removed from $y b$ - So, the election configuration of ${y} b^{2+}$ is $[x e] \& f^{14}$. The configuration $\left[X_e\right] 4 f^{14}$ has no unpaired electrons. All the electrons are paired and hence $yb^{2+}$ is diamagnetic. Element Lu electron configuration is $\left[x_e\right] 4 f^{\prime 4} 5 d^1 6 s^2$ For $\mathrm{Lu}^{3+}$, three valence electrons we removed from. Lu . So, the electron configuration of $\mathrm{Lu}^{3+}$ is $\left[x_e\right] 4 f^{14}$. The configuration $\left[\mathrm{Xe}\right] 4 f^{14}$ has no unpaired electrons. All the electrons are paired and hence $L u^3 t$ is diamagnetic. So, It is a pair of diamagnetic ions: c) $\mathrm{La}^{2+}, \mathrm{Ce}^{3+}$ Element la electron configuration is [ $x e$ ] $5 d^{\prime} 6 s^2$. For $\mathrm{La}^{2+}$, two valence electrons are removed so, the electron configuration of $\mathrm{La}^{2+}$ is $[\mathrm{Xe}] 5 d^{\prime}$. The configuration $\left[X_e\right] 5 d^{\prime}$ has one unpaired election and hence it is paramagnetic, not diamagnetic. Element $C_e$ electron configuration is $[x e] 4 f^{\prime} 5 d^{\prime} 6 s^2$ For $\mathrm{Ce}^{3+}$, three valence electrons we removed. So, the electron configuration of $\mathrm{Ce}^{3+}$ is $[x e] 4 f^{\prime}$ The configuration $[\mathrm{Xe}]$. $4f^{\prime}$ has one unpaired electron and hence it is paramagnetic, not diamagnetic. So, it is not a pair of diamagnetic ions. D) $\mathrm{Yb}^{3+}, \mathrm{Lu}^{2+}$ yb electron configuration is $[x e] 4 f^{14} 6 s^2$. For $y b^{3+}$, three valence electrons are removed from $y b$. So, the electron configuration of $\mathrm{Yb}^{3+}$ is $\left[x_e\right] 4 f^{13}$ The configuration $\left[x_e\right] 4 f^{13}$ has unpaired election and it is paramagnetic, not diamagnetic. Lu electron configuration is $\left[X_e\right] 4 f^{14} 5 d^1 6 s^2$. For $L u^{2+}$, two valence electrons are removed from LU. So, the electron configuration of $\mathrm{Lu}^{2+}$ is $\left[\mathrm{Xe}_e\right] 4 f^{14} 5 d^1$. The configuration $[x]$. $4 f^{\prime 4} 5 d^{\prime}$ has unpaired electron and it is paramagnetic, not diamagnetic. So, it is not a pair of diamagnetic ions. The pairs of diamagnetic ions are A) $\mathrm{La}^{3+}, \mathrm{Cl}^{4+}$ B) $\mathrm{Yb}^{2+}, \mathrm{Lu}^{3+}$

$\mathrm{PbS+dil.\,HNO_3\to Pb(NO_3)_2+S+NO+H_2O}$

NO is paramagnetic due to the presence of unpaired electron. It is a neutral oxide. It is colourless.

Hence, (A) and (D) are correct statements.

The explanation of given statements are as follows :

(a) Aqua-regia is prepared by mixing conc. HCl and conc. HNO₃ in 3 : 1 (v/v) ratio and is used in oxidation of gold and platinum. Hence, option (a) is correct.

(b) Yellow colour of aqua-regia is due to its decomposition into NOCl (orange yellow) and Cl₂ (greenish yellow). Hence, option (b) is correct.

(c) When gold reacts with aqua-regia then it produces $AuCl_4^ - $ anion complex in which Au has +3 oxidation state.



Hence, option (c) is correct.

(d) Reaction of gold with aqua-regia produces NO gas in absence of air. Hence, option (d) is incorrect.

$MnO_4^{2 - }$ ion has one unpaired electrons, therefore it gives d-d transition to form green colour. Y complex has paramagnetic nature due to presence of one unpaired electron.

In aqueous solution,



$MnO_4^{2 - }$ ions gives charge transfer spectrum in which a fraction of electronic charge is transferred between the molecular entities.

$ \because $ $\mathop {MnO_4^{2 - }}\limits_{(Y)} \mathrel{\mathop{\kern0pt\longrightarrow} \limits_{oxidation}^{Electrolytic}} \mathop {MnO_4^ - }\limits_{(Z)} + {e^ - }$

In acidic medium, Y undergoes disproportionation reaction.



$\mathop {MnO_4^{2 - }}\limits_{(Y)} $ and $\mathop {MnO_4^ - }\limits_{(Z)} $ both ions form $\pi $-bonding between p-orbitals of oxygen and d-orbitals of manganese.

Thus, options (a, c, d) are correct.

Statement wise explanation is

Statement (a) Mn²⁺ produces yellow-green colour in flame test while Cu²⁺ produces bluish-green colour in flame test. Thus, due to the presence of green colour in both the cases, flame test is not the suitable method to distinguish between nitrate salts of Cu²⁺ and Mn²⁺. Hence this statement is wrong.

Statement (b) Cu²⁺ belong to group II of cationic or basic radicals. It gives black ppt. of CuS if H₂S is passed through it in the presence of acid (e.g. HCl). Mn₂₊ does not show this property hence this can be considered as a suitable method to distinguish between Mn²⁺ and Cu²⁺. Hence, this statement is correct.

Statement (c) In faintly basic medium when H₂S is passed both Cu²⁺ and Mn²⁺ forms precipitates. Thus, it is not suitable method to distinguish between them. Hence, this statement is incorrect.

Statement (d) The standard reduction potential of Cu²⁺ /Cu is +0.34 V while that of Mn²⁺/Mn is $-$1.18 V. This can be used to distinguish between Cu²⁺ and Mn²⁺. In general less electropositive metals have higher SRP. Hence, this statement is correct.

Compounds which dissolve in both acids and bases are known as amphoteric oxides.

Amphoteric oxides are :

Cr₂O₃, BeO, SnO, SnO₂, ZnO, Al₂O₃ , PbO and PbO₂ Whereas, NO is a neutral oxide, B₂O₃ is an acidic oxide and CrO is a basic oxide.

Consider the reactions of the given compounds with $\mathrm{S}^{2-}$ and $\mathrm{SO}_4^{2-}$.

(i) With $\mathrm{CuCl}_2$ :

$ \begin{aligned} &\mathrm{CuCl}_2+\mathrm{S}^{2-} \rightarrow \underset{\text { black ppt }}{\mathrm{CuS}}+2 \mathrm{Cl}^{-} \\\\ &\mathrm{CuCl}_2+\mathrm{SO}_4{ }^{2-} \rightarrow \text { No precipitate } \end{aligned} $

(ii) With $\mathrm{BaCl}_2$ : Alkaline earth metal sulphides are sparingly soluble in water but pass into solution with time and hence do not give precipitates from their aqueous solutions.

$ \begin{aligned} &\mathrm{BaCl}_2+\mathrm{S}^{2-} \rightarrow \underset{\text { No ppt }}{\mathrm{BaS}}+2 \mathrm{Cl}^{-} \\\\ &\mathrm{BaS}+2 \mathrm{H}_2 \mathrm{O} \rightleftharpoons \underset{\text { water soluble }}{\mathrm{Ba}(\mathrm{OH})_2+{\mathrm{Ba}(\mathrm{SH})_2}} \end{aligned} $

Alkaline earth metal sulphates, $\mathrm{BaSO}_4, \mathrm{CaSO}_4$ and $\mathrm{SrSO}_4$ are insoluble in water.

$\mathrm{BaCl}_2+\mathrm{SO}_4^{2-} \rightarrow \underset{\text { white ppt }}{\mathrm{BaSO}_4}+2 \mathrm{Cl}^{-}$

(iii) With $\mathrm{Pb}(\mathrm{OAc})_2$ :

$ \begin{aligned} &\mathrm{Pb}(\mathrm{OAc})_2+\mathrm{S}^{2-} \rightarrow \underset{\text { black ppt }}{\mathrm{PbS}}+2 \mathrm{CH}_3 \mathrm{COO}^{-} \\\\ &\mathrm{Pb}(\mathrm{OAc})_2+\mathrm{SO}_4^{2-} \rightarrow \underset{\text { white ppt }}{\mathrm{PbSO}_4}+2 \mathrm{CH}_3 \mathrm{COO}^{-} \end{aligned} $

The $K_{\mathrm{sp}}$ of $\mathrm{PbS}$ is $3 \times 10^{-28}$ and of $\mathrm{PbSO}_4$ is $25 \times$ $10^{-28}$. The difference is large, thus only $\mathrm{PbS}$ is selectively precipitated.

(iv) With $\mathrm{Na}_2\left[\mathrm{Fe}(\mathrm{CN})_5 \mathrm{NO}\right]$ : When sodium nitroprusside is added to solution of sulphide ions, purple colouration is observed. This is a confirmative test for sulphides.

$ \begin{aligned} &{\left[\mathrm{Fe}(\mathrm{CN})_5 \mathrm{NO}\right]^{2-}+\mathrm{S}^{2-} \rightarrow \underset{\text { Purple colour }}{\left[\mathrm{Fe}(\mathrm{CN})_5 \mathrm{NOS}\right]}} \\\\ &\mathrm{Na}_2\left[\mathrm{Fe}(\mathrm{CN})_5 \mathrm{NO}\right]+\mathrm{SO}_4{ }^{2-} \rightarrow \text { No precipitate } \end{aligned} $

The pairs Cu²⁺, Pb²⁺ and Hg²⁺, Bi³⁺ are precipitated as sulphides upon passing H₂S in acidic medium.

The correct statements are as follows:

(A) Cr³⁺ is more stable than Cr²⁺, thus, Cr²⁺ act as reducing agent.

(B) Mn²⁺ is more stable than Mn³⁺, thus, Mn³⁺ act as an oxidizing agent.

(C) Both Cr²⁺ and Mn³⁺ exhibit d⁴ electronic configuration.

To understand which of the given reagents will give copper metal upon heating with $Cu_2S$, let's discuss each one based on their chemical reactions during the process:

Option A: CuFeS₂

$CuFeS_2$ (Chalcopyrite) when heated with $Cu_2S$, it can react to form copper metal as part of the copper extraction process. However, the direct reaction between $CuFeS_2$ and $Cu_2S$ is more involved and typically requires a series of steps, including smelting and conversion to get to the copper metal. Therefore, while $CuFeS_2$ is a source of copper, it's not directly because of its reaction with $Cu_2S$, but through a complex metallurgical process.

Option B: CuO

$ \begin{aligned} & 4 \mathrm{CuO} \xrightarrow{\Delta} 2 \mathrm{Cu}_2 \mathrm{O}+\mathrm{O}_2 \\\\ & \mathrm{Cu}_2 \mathrm{~S}+2 \mathrm{Cu}_2 \mathrm{O} \xrightarrow{\Delta} 6 \mathrm{Cu}+\mathrm{SO}_2 \end{aligned} $

Option C: Cu₂O

Copper(I) oxide ($Cu_2O$) can also react with $Cu_2S$ to produce copper metal directly. The equation for this reaction is:

$2Cu_2O + Cu_2S \rightarrow 6Cu + SO_2$

Here, $Cu_2O$ serves as the oxidizing agent, reducing $Cu_2S$ to copper metal, while itself being reduced to copper, and sulfur is oxidized to sulfur dioxide. This reaction is similar in principle to the reaction with $CuO$, but involves copper(I) oxide instead.

Option D: CuSO₄

$\begin{array}{r}\mathrm{CuSO}_4 \xrightarrow{720^{\circ} \mathrm{C}} \mathrm{CuO}+\mathrm{SO}_2+\frac{1}{2} \mathrm{O}_2 \\\\ 4 \mathrm{CuO} \xrightarrow{1100^{\circ} \mathrm{C}} 2 \mathrm{Cu}_2 \mathrm{O}+\mathrm{O}_2 \\\\ 2 \mathrm{Cu}_2 \mathrm{O}+\mathrm{Cu}_2 \mathrm{~S} \xrightarrow{\Delta} 6 \mathrm{Cu}+\mathrm{SO}_2\end{array}$

Reaction of alkali metals with ammonia depends upon the physical state of ammonia whether it is in gaseous state or liquid state. If ammonia is considered as a gas then reaction will be

(a) $Na + \mathop {N{H_3}}\limits_{(Excess)} \to NaN{H_2} + {1 \over 2}{H_2}$

(NaNH₂ + 1/2H₂ are diamagnetic) If ammonia is considered as a liquid then reaction will be

$M + (x + y)N{H_3} \to {[M{(N{H_3})_x}]^ + } + {[e{(N{H_3})_y}]^ - }$

$\bullet$ ammoniated electron

$\bullet$ blue colour

$\bullet$ paramagnetic

$\bullet$ very strong reducing agent

(b) $K + \mathop {{O_2}}\limits_{(Excess)} \to \mathop {K{O_2}({K^ + },O_2^ - )}\limits_{Potassium\,\sup eroxide\,Paramagnetic} $

(c) $3Cu + 8HN{O_3} \to \mathop {3Cu{{(N{O_3})}_2}}\limits_{Paramagnetic} + \mathop {2NO}\limits_{Paramagnetic} + 4{H_2}O$

(d)
Hence, option (a), (b) and (c) are correct choices.

The first reaction is

${K_3}[Fe{(CN)_6}] + \mathop {KI}\limits_{excess} \to {K_4}[Fe{(CN)_6}] + \mathop {K{I_3}}\limits_{brownish\,yellow\,solution} $

This is a redox reaction as both oxidation and reduction are taking place.

$\mathop {{K_4}[Fe{{(CN)}_6}]}\limits_{soluble} \, + ZnS{O_4} \to \mathop {{K_2}Z{n_3}{{[Fe{{(CN)}_6}]}_3}}\limits_{white\,ppt} + NaOH \to \mathop {N{a_2}{{[Zn{{(OH)}_4}]}^{2 - }}}\limits_{soluble} $

$2I_3^ - + 2N{a_2}{S_2}{O_3} \to \mathop {N{a_2}{S_4}{O_6}}\limits_{clear\,solution} + 2NaI + \mathop {2{I_2}}\limits_{(blue\,colour)} $

$ \mathrm{MnCl}_2 \stackrel{\mathrm{NaOH}(\mathrm{aq})}{\longrightarrow} \underset{(\mathrm{P})}{\mathrm{Mn}\left(\mathrm{OH}_2\right)}+\underset{(\mathrm{Q})}{\mathrm{NaCl}} $

$ \mathrm{Mn}(\mathrm{OH})_2 \underset{\mathrm{PbO}_2(\text { excess })}{\stackrel{\text { aq. } \mathrm{H}_2 \mathrm{SO}_4}{\longrightarrow}} \underset{(\mathrm{X})}{\mathrm{MnO}_4^{-}}+\mathrm{Pb}^{2+} $

$ \mathrm{NaCl} \frac{\mathrm{MnO}(\mathrm{OH})_2}{\text { conc. } \mathrm{H}_2 \mathrm{SO}_4 \text { warm }} \underset{(\mathrm{Y})}{\mathrm{Cl}_2} $

Borax bead test is performed only for coloured salt. Borax (sodium pyroborate), Na₂B₄O₇.10H₂O on heating gets fused and lose water of crystallisation. It swells up into fluffy white porous mass which melts into a colourless liquid which later form a clear transparent glassy bead consisting of boric anhydride and sodium metaborate.



Boric anhydride is non-volatile. When it react with Cr(III) salt then deep green complex is formed.



Hence, option (c) is correct.

(a) Conc. HNO₃ makes iron passive. Cold relatively concentrated HNO₃ will react with Fe.

$\mathop {Fe}\limits_{Iron} + \mathop {6HN{O_3}}\limits_{Nitric\,acid} \to \mathop {Fe{{(N{O_3})}_3}}\limits_{Iron\,nitrate} + \mathop {3N{O_2}}\limits_{Nitrogen\,dioxide} + \mathop {3{H_2}O}\limits_{Water} $

(b) $\mathop {Cu}\limits_{Copper} + \mathop {4HN{O_3}(conc.)}\limits_{Nitric\,acid} \to \mathop {Cu{{(N{O_3})}_2}}\limits_{Copper\,nitrate} + \mathop {2N{O_2}}\limits_{Nitrogen\,dioxide} + \mathop {2{H_2}O}\limits_{Water} $

(c) $\mathop {Zn}\limits_{Zinc} + \mathop {2NaO{H_{(aq)}}}\limits_{Sodium\,hydroxide} \to \mathop {N{a_2}Zn{O_2}}\limits_{Sodium\,zincate} + \mathop {{H_2}}\limits_{Hydrogen} $

(d) $\mathop {4Au}\limits_{Gold} + \mathop {8NaCN}\limits_{Sodium\,cyanide} + \mathop {{O_2}}\limits_{Oxygen} + \mathop {2{H_2}O}\limits_{Water} \to \mathop {4Na[Au{{(CN)}_2}]}\limits_{Sodium\,dicyanoaurate} + \mathop {4NaOH}\limits_{Sodium\,hydroxide} $

$A{g^ + }\mathop {2{S_2}O_3^{2 - }}\limits_{(Excess)} \to \mathop {{{[Ag{{({S_2}{O_3})}_2}]}^{3 - }}}\limits_{Argentothiosulphate\,(X)\,(Soluble)} \buildrel {A{g^ + }} \over \longrightarrow \mathop {A{g_2}{S_2}{O_3}}\limits_{(Y)\,(White\,ppt.)} \mathrel{\mathop{\kern0pt\longrightarrow} \limits_{(due\,to\,hydrolysis)}^{With\,time}} \mathop {A{g_2}S}\limits_{(Z)\,(Black\,ppt.)} $

On treatment with H₂S, Zn(II) gets precipitated as ZnS. Mg(II) does not get precipitated, while Fe³⁺ and Al³⁺ get precipitated as hydroxides.

When we consider the color of light absorbed by a substance, we are referring to the complementary color of the light that is transmitted or reflected by the substance. An aqueous solution of copper(II) sulfate (CuSO₄) appears blue to our eyes. This color results from the absorption of light in the complementary color range.

The visible spectrum of light can be divided into several regions, each corresponding to a different color. The complementary color of blue is orange-red. This means that the aqueous solution of CuSO₄ absorbs light in the orange-red region of the spectrum.

Therefore, the correct answer is:

Option A: orange-red